ostrosłup prawidłowy czworokątny - w podstawie kwadrat
AB=BD=DC=AC=a
kąt CFB=90st
AF=DF=CF=BF=b=5\sqrt{2}
FE=H
V=\frac{1}{3}P_p*H=a^2*H
CB=d (przekątna kwdratu)
d^2=(CF)^2+(BF)^2=b^2+b^2=2b^2=2*(5\sqrt{2})^2=2*25*2=100
d=\sqrt{100}=10
d=a\sqrt{2}
a\sqrt{2}=10
a=\frac{10}{\sqrt{2}}=\frac{10\sqrt{2}}{\sqrt{2}*\sqrt{2}}=\frac{10\sqrt{2}}{2}=5\sqrt{2}
H^2=b^2-(\frac{1}{2}d)^2=(5\sqrt{2})^2-5^2=25*2-25=50-25=25
H=\sqrt{25}=5
V=a^2*H=(5\sqrt{2})^2*5=25*2*5=250